Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^{2}(a+b)}$.
But I am not getting the answer.
 A: The case $a^2=b^2$ being simple, let's just consider, by symmetry, the case $a>b>0$.
Observe that
$$
\partial _x \left(\frac{1}{a^2 \cos^2x+b^2 \sin^2 x}\right)=2(a^2-b^2)\frac{\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 x)^2}
$$
then, integrating by parts, you may write
$$
\begin{align}
I(a,b)&=\int_0^{\pi/2}\frac{x\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 x)^2} dx\\\\
&=\frac{x}{2(a^2-b^2)}\left.\frac{1}{a^2 \cos^2x+b^2 \sin^2 x}\right|_{0}^{\pi/2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{dx}{a^2 \cos^2x+b^2 \sin^2 x}\\\\
&= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{1}{\left(a^2 +b^2 \large{\frac{\sin^2 x}{\cos^2x}}\right)}\frac{1}{\cos^2 x}dx\\\\
&= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\int_0^{\pi/2}\frac{1}{\left(a^2 +b^2 \tan ^2x\right)}(\tan x)'dx\\\\
&= \frac{\pi}{4(a^2-b^2)b^2}-\frac{1}{2(a^2-b^2)}\frac{1}{ab}\left.\arctan \left(\frac ba \tan x \right)\right|_{0}^{\pi/2}\\\\
&= \frac{\pi}{4(a^2-b^2)b^2}-\frac{\pi}{4(a^2-b^2)}\frac{1}{ab}\\\\
&= \frac{\pi}{4(a+b)ab^2}.
\end{align}
$$ 
A: let $x=\dfrac{t}{2}$, we have
$$I=\dfrac{1}{2}\int_{0}^{\pi}\dfrac{\dfrac{t}{2}\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}}{\left(a^2\sin^2{\dfrac{t}{2}}+b^2\cos^2{\dfrac{t}{2}}\right)^2}dt=\dfrac{1}{2}\int_{0}^{\pi}\dfrac{t\sin{t}}{[(a^2+b^2)+(a^2-b^2)\cos{t}]^2}dt$$
So
\begin{align*}I&=-\dfrac{1}{2(a^2-b^2)}\int_{0}^{\pi}t\;d\left(\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}\right)\\
&=-\dfrac{1}{2(a^2-b^2)}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}\Bigg|_{0}^{\pi}+\int_{0}^{\pi}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}dt\\
&=-\dfrac{\pi}{4b^2(a^2-b^2)}+\int_{0}^{\pi}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}dt
\end{align*}
A: Integrating by parts,
$$\int\frac{x\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
$$=x\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx-\int\left[\frac{dx}{dx}\int\frac{\sin x\cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx\right]dx$$
Now $a^2\cos^2x+b^2\sin^2x=u\implies2(b^2-a^2)\sin x\cos x\ dx=du$
For $\displaystyle\int\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\int\frac{\sec^2x\ dx}{a^2+b^2\tan^2x},$
set $b\tan x=a\tan y$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{0}^{\pi/2}{x\sin\pars{x}\cos\pars{x}\over
\bracks{a^{2}\cos^{2}\pars{x} + b^{2}\sin^{2}\pars{x}}^{2}}\,\dd x}
\\[5mm]&=\ \overbrace{\int_{0}^{\pi/2}{x\sin\pars{2x}/2\over\braces{%
a^{2}\bracks{1 + \cos\pars{2x}}/2 + b^{2}\bracks{1 - \cos\pars{2x}}/2}^{2}}\,\dd x}
^{\dsc{2x}\ \ds{\mapsto}\ \dsc{x}}
\\[5mm]&=\half\int_{0}^{\pi}{x\sin\pars{x}\over\bracks{%
a^{2} + b^{2} + \pars{a^{2} - b^{2}}\cos\pars{x}}^{2}}\,\dd x
\\[5mm]&={1 \over 2\pars{a^{2} - b^{2}}}\int_{x\ =\ 0}^{x\ =\ \pi}x\,
\dd\bracks{1 \over a^{2} + b^{2} + \pars{a^{2} - b^{2}}\cos\pars{x}}
\\[1cm]&={1 \over 2\pars{a^{2} - b^{2}}}\,\left.
{x \over a^{2} + b^{2} + \pars{a^{2} - b^{2}}\cos\pars{x}}
\right\vert_{x\ =\ 0}^{x\ =\ \pi}
\\[5mm]&-{1 \over 2\pars{a^{2} - b^{2}}}\ \underbrace{\int_{0}^{\pi}
{\dd x \over a^{2} + b^{2} + \pars{a^{2} - b^{2}}\cos\pars{x}}}
_{\dsc{t}\ \ds{=}\ \dsc{\tan\pars{x \over 2}}}
\\[1cm]&={\pi \over 4b^{2}\pars{a^{2} - b^{2}}}
-{1 \over 2\pars{a^{2} - b^{2}}}\ \overbrace{\int_{0}^{\infty}
{\dd x \over b^{2}t^{2} + a^{2}}}^{\ds{=}\ \dsc{\pi \over 2\verts{ab}}}
={\pi \over 4\verts{b}\pars{a^{2} - b^{2}}}
\pars{{1 \over \verts{b}} - {1 \over \verts{a}}}
\\[5mm]&=\color{#66f}{\large%
{\pi \over 4\verts{a}b^{2}\pars{\verts{a} + \verts{b}}}}
\end{align}
A: You can use this property : $$\int_a^b f(x)\hspace{1mm}dx = \int_a^b f(a+b-x)\hspace{1mm}dx$$
To prove this property : Substitute $a+b-x = u$
Let us have $$I = \int_0^{\pi/2}\dfrac{x\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (1)$$
After applying the property, you will get  
$$I = \int_0^{\pi/2}\dfrac{(\pi/2-x)\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (2)$$
After adding the two equation, you get
$$I =\dfrac{\pi}{4} \int_0^{\pi/2}\dfrac{\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx $$
Now you will substitute $a^2\cos^2x+b^2\sin^2x = u$
To get $$I = \dfrac{\pi}{8(b^2-a^2)}\int_{a^2}^{b^2} \dfrac{1}{u^2}\hspace{1mm}du$$
I hope you can Integrate from here
